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Question: If the coefficients of \(2^{n}\) terms of \((1 + x - 3x^{2})^{3148}\) are in A.P., then r =...

If the coefficients of 2n2^{n} terms of (1+x−3x2)3148(1 + x - 3x^{2})^{3148} are in A.P., then r =

A

6

B

7

C

8

D

9

Answer

9

Explanation

Solution

0<(2−1)<10 < (\sqrt{2} - 1) < 1; k+f+f′=(2+1)6+(2−1)6k + f + f^{'} = (\sqrt{2} + 1)^{6} + (\sqrt{2} - 1)^{6}

By the given condition

=2{6C0.23+6C2.22+6C4.2+6C6}=198= 2\left\{ 6C_{0}.2^{3} +^{6}C_{2}.2^{2} +^{6}C_{4}.2 +^{6}C_{6} \right\} = 198 …..(i)

⇒ ∴f+f′=198−k=\therefore f + f^{'} = 198 - k =

⇒ 0≤f<10 \leq f < 1

=0<f′<1⇒0<(f+f′)<20 < f^{'} < 1 \Rightarrow 0 < (f + f^{'}) < 2+⇒f+f′=1\Rightarrow f + f^{'} = 1

⇒(∵f+f′(\because f + f^{'}

⇒ f+f′f + f^{'}

⇒ (n+1)(n+2)2=45\frac{(n + 1)(n + 2)}{2} = 45

⇒ n2+3n−88=0⇒n=8n^{2} + 3n - 88 = 0 \Rightarrow n = 8

⇒ ∵\because

⇒ a3+b3+3ab(a+b)=(a+b)3a^{3} + b^{3} + 3ab(a + b) = (a + b)^{3}

⇒ ∴Nr=(18+7)3=253\therefore N^{r} = (18 + 7)^{3} = 25^{3}

But 5 is not given. Hence r = 9.

Trick : Put the value of r from options in equation (i), only (4) satisfy it.